###### Question 17360 – NTA UGC NET Dec 2023 Paper-2

May 23, 2024###### Question 17364 – NTA UGC NET Dec 2023 Paper-2

May 23, 2024# NTA UGC NET Dec 2023 Paper-2

Question 22 |

(A), (B) and (C) Only | |

(B) and (C) Only | |

(C) (D) and (E) Only | |

(C) and (E) Only | |

∀n∃m |

(B) ∃(n) ∀(m) (n < m^2): This statement is also false. It claims that there exists an integer n such that for all integers m, n is less than m^2. However, this is not possible because for any integer n, there will always be an integer m such that m^2 is less than or equal to n.

(C) ∃(n) ∀(m) (nm = m): This statement is true. It claims that there exists an integer n such that for all integers m, the product of n and m is equal to m. This is true for the integer n = 1, as 1 multiplied by any integer m equals m.

(D) ∃(n) ∃(m) (n^2 + m^2 = 6): This statement is false. It claims that there exist integers n and m such that the sum of their squares is 6. However, the only way to get a sum of 6 from two squares is 1^2 + 1^2 = 2 or 2^2 + 2^2 = 8, neither of which equals 6.

(E) ∃(n) ∃(m) (n + m=4 AND n-m=1): This statement is true. It claims that there exist integers n and m such that their sum is 4 and their difference is 1. This is true for the integers n = 2.5 and m = 1.5, as 2.5 + 1.5 = 4 and 2.5 – 1.5 = 1.

(B) ∃(n) ∀(m) (n < m^2): This statement is also false. It claims that there exists an integer n such that for all integers m, n is less than m^2. However, this is not possible because for any integer n, there will always be an integer m such that m^2 is less than or equal to n.

(C) ∃(n) ∀(m) (nm = m): This statement is true. It claims that there exists an integer n such that for all integers m, the product of n and m is equal to m. This is true for the integer n = 1, as 1 multiplied by any integer m equals m.

(D) ∃(n) ∃(m) (n^2 + m^2 = 6): This statement is false. It claims that there exist integers n and m such that the sum of their squares is 6. However, the only way to get a sum of 6 from two squares is 1^2 + 1^2 = 2 or 2^2 + 2^2 = 8, neither of which equals 6.

(E) ∃(n) ∃(m) (n + m=4 AND n-m=1): This statement is true. It claims that there exist integers n and m such that their sum is 4 and their difference is 1. This is true for the integers n = 2.5 and m = 1.5, as 2.5 + 1.5 = 4 and 2.5 – 1.5 = 1.